CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
years012€12 000FV = ?

You invest €12 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 2 years.

PV = €12 0004.8% / 4n = 2
FV = €
2.Time value of moneyHard
months0123…36€25 000A?AAA

A loan of €25 000 is repaid with equal monthly payments over 3 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.

P = €25 0002.4% / 12N = 36
A = €
3.NPV & IRRMedium
years01234−30 00011 0009 0009 00012 000

An investment of €30 000 gives these cash flows at the end of years 1 to 4: €11 000, €9 000, €9 000, €12 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −30 000CF1 = 11 000CF2 = 9 000CF3 = 9 000CF4 = 12 000
NPV = €
4.NPV & IRRHard
years012−20 00010 50012 000

A project costs €20 000 and returns €10 500, €12 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 10 500CF2 = 12 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 3.5%n = 9
P = €
6.Bond pricing & yieldHard
years01234€1 162€60€60€60 + €1 060

A bond with 10 years to maturity (face value €1 000, annual coupon 6%) sells for €1 162. The issuer can call it in 4 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 162call price = €1 060call in 4 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 10%debt€5 000 000rd​ = 4.5%T = 21%

A company has equity worth €10 000 000 and debt worth €5 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.

E = €10 000 000D = €5 000 000re = 10%rd = 4.5%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €34.00β = 1.1rf​ = 3.5%, rm​ = 9%debt20 000 × €1 050rd​ = 5%T = 28%

A company has 5 000 000 shares at €34.00 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 9%. Find the WACC.

β = 1.1rf = 3.5%rm = 9%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 14, 60, 33, 21, 52.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 56.

μ = 60σ = 5x = 56
z =
11.ProbabilityMedium
xp(x)00.3510.120.3530.0540.15μ = 1.55

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3, 4p(x): 0.35, 0.1, 0.35, 0.05, 0.15
Var(X) =
12.ProbabilityHard
n = 9, p = 0.5: P(X = 3) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?

n = 9p = 0.5k = 3
P =
13.Normal distributionMedium
32 min30.8 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30.8 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30.8 min
P =
14.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h
15.Confidence intervalsMedium
n = 15, s = 18L = ?Ux̄ = 116E90%

A sample of 15 items has mean 116 and standard deviation s = 18. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 15x̄ = 116s = 1890%t(14) = 1.761
lower limit =
16.Confidence intervalsHard
σ = 35, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 35. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?

σ = 35E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

Two teaching methods are compared. Group 1: 8 students, mean score 72, standard deviation 9. Group 2: 6 students, mean score 59.7, standard deviation 12. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 8x̄₁ = 72s₁ = 9n₂ = 6x̄₂ = 59.7s₂ = 12
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 12, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 12. Find the minimum sample size n.

α = 0.05power = 95%δ = 4σ = 12zα = 1.645zβ = 1.645
n =
19.Linear regression & correlationMedium
xy137243343450552xy24404550

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 225Σx² = 55Σy² = 10271Σxy = 712
r =
20.Linear regression & correlationHard
xy264126198261030xy510102030x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 26) is made. Find its residual e = y − ŷ.

n = 5(9, 26)
e =

3Solutions

1.Time value of money

Answer: FV = 13202 €

Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.

Rate and number of periods

i = r/m = 0.048/4 = 0.012
N = m·n = 4·2 = 8

Future value

FV = 12 000·(1 + 0.012)8 = 12 000·1.1 = €13 201.56

Answer

FV = €13 201.56
2.Time value of money

Answer: A = 720.4 €

Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.

Monthly rate and number of payments

i = 0.024/12 = 0.002
N = 12·3 = 36

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.9306
A = 25 000·0.002/(1 − 0.9306) = €720.44

Answer

A = €720.44
3.NPV & IRR

Answer: NPV = 6286 €

Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.

Present value of each cash flow

CF0 = −30 000
CF1/(1 + 0.05)1 = 11 000/1.05 = 10 476.19
CF2/(1 + 0.05)2 = 9 000/1.103 = 8 163.27
CF3/(1 + 0.05)3 = 9 000/1.158 = 7 774.54
CF4/(1 + 0.05)4 = 12 000/1.216 = 9 872.43

Add them up

NPV = €6 286.42

Answer

NPV = €6 286.42

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 8.037 %

Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.

Bracket the IRR

NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.

NPV(5%) = 884.35
NPV(10%) = −537.19

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 5 + 5·884.35/(884.35 − (−537.19)) = 8.111%
exact: NPV(8.037%) = 0 → IRR = 8.037%

Answer

IRR = 8.037%
5.Bond pricing & yield

Answer: P = 1268 €

Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.

Per period

C = €35.00, i = 0.0175, N = 18

Present value of the coupons

€35.00·(1 − 0.7318)/0.0175 = €536.44

Present value of the face value

€1 000·0.7318 = €731.78

Price

P = €1 268.22

Answer

P = €1 268.22

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTC = 3.067 %

Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.

Equation for the yield

1162 = 60·(1 − (1 + y)−4)/y + 1060/(1 + y)4

Solve numerically

y = 3%: €1 164.82
y = 4%: €1 123.89
YTC = 3.067%

Answer

YTC = 3.067%

The yield to maturity is 4.002%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: WACC = 7.852 %

Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).

Weights

V = €15 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·10% + 0.3333·4.5%·(1 − 0.21) = 7.852%

Answer

WACC = 7.852%
8.CAPM & WACC

Answer: WACC = 8.896 %

Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).

Market values

E = 5 000 000·€34.00 = €170 000 000
D = 20 000·€1 050 = €21 000 000

Cost of equity (CAPM)

re = 3.5% + 1.1·5.5% = 9.55%

Weights

V = €191 000 000, E/V = 0.8901, D/V = 0.1099

WACC

WACC = 0.8901·9.55% + 0.1099·5%·(1 − 0.28) = 8.896%

Answer

WACC = 8.896%
9.Descriptive statistics

Answer: s = 19.69

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 180/5 = 36

Squared deviations

Σ(x − x̄)² = (−22)² + 24² + (−3)² + (−15)² + 16² = 1550

Standard deviation

s = √(1550/4) = 19.69

Answer

s = 19.69
10.Descriptive statistics

Answer: z = −0.8

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (56 − 60)/5 = −0.8

Answer

z = −0.8
11.Probability

Answer: Var(X) = 1.948

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.35 + 1·0.1 + 2·0.35 + 3·0.05 + 4·0.15 = 1.55

Mean of the squares

E(X²) = 0·0.35 + 1·0.1 + 4·0.35 + 9·0.05 + 16·0.15 = 4.35

Variance

Var(X) = 4.35 − 1.55² = 1.948

Answer

Var(X) = 1.948
12.Probability

Answer: P = 0.1641

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 3) = 84·0.53·0.56 = 84·0.125·0.01563 = 0.1641

Answer

P = 0.1641
13.Normal distribution

Answer: P = 0.6844

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (30.8 − 32)/2.5 = −0.48

Upper tail

P = 1 − Φ(−0.48) = 1 − 0.3156 = 0.6844

Answer

P = 0.6844
14.Normal distribution

Answer: x = 578.4 h

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h
15.Confidence intervals

Answer: lower limit = 107.8

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Critical value

df = n − 1 = 14, t = 1.761

Margin of error

E = 1.761·18/√15 = 8.184

Interval

116 ± 8.184 → [107.8, 124.2]

Answer

lower limit = 107.8
16.Confidence intervals

Answer: n = 523

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (1.96·35/3)² = 522.9

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 523

Answer

n = 523
17.Hypothesis tests & confidence intervals

Answer: t = 2.199

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Pooled variance

sp² = (7·9² + 5·12²)/12 = 107.3
sp = 10.36

Standard error

SE = 10.36·√(1/8 + 1/6) = 5.593

Test statistic

t = (72 − 59.7)/5.593 = 2.199

Answer

t = 2.199

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 98

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·12/4)² = 97.42

Round up

n = 98

Answer

n = 98
19.Linear regression & correlation

Answer: r = 0.9683

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Centred sums

Sxy = 712 − 15·225/5 = 37
Sxx = 55 − 15²/5 = 10
Syy = 10271 − 225²/5 = 146

Correlation coefficient

r = 37/√(10·146) = 0.9683

Answer

r = 0.9683
20.Linear regression & correlation

Answer: e = −1.9

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 30/5 = 6
ȳ = 93/5 = 18.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124

Regression line

b = 124/40 = 3.1
a = 18.6 − 3.1·6 = 0
ŷ = 0 + 3.1·x

Predicted value

ŷ = 0 + 3.1·9 = 27.9

Residual

e = 26 − 27.9 = −1.9

Answer

e = −1.9

The point lies below the line.

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